Variance of a weighted population

[rogo0034]

Homework Statement

Homework Equations

The Attempt at a Solution

Was able to get the population mean of 5.3 with 4*.2 + 5*.4 + 6*.3 + 7*.1
But for some reason I'm drawing blank for the variance (which i guess is .81)

 

[RoshanBBQ]

Homework Statement

Homework Equations

The Attempt at a Solution

Was able to get the population mean of 5.3 with 4*.2 + 5*.4 + 6*.3 + 7*.1
But for some reason I'm drawing blank for the variance (which i guess is .81)

Variance is defined as:
var\{x\}=E\{(x-E\{x\})^2\}
So if you wanted to directly apply the equation above, you would compute it the same way as mean, except instead of doing a weighted sum of x, you would do a weighted sum of (x-E{x})^2. So for each x, before doing the weighted sum, you would subtract off the mean you found already and square it.

But there is a popular reworking of the equation:
var\{x\}=E\{x^2-2xE\{x\}+E^2\{x\}\}=E\{x^2\}-E^2\{x\}

So in this case, you would compute the expected value of x^2 (the same way you did the mean except use x^2 instead of x). Then, subtract the square of the expectation you already found.

 

[rogo0034]

Got it! thanks, that was easier